Problem: Simplify; express your answer in exponential form. Assume $n\neq 0, x\neq 0$. $\dfrac{{(n)^{-1}}}{{(n^{-4}x^{3})^{-1}}}$
Explanation: To start, try working on the numerator and the denominator independently. In the numerator, we have ${n}$ to the exponent ${-1}$ . Now ${1 \times -1 = -1}$ , so ${(n)^{-1} = n^{-1}}$ In the denominator, we can use the distributive property of exponents. ${(n^{-4}x^{3})^{-1} = (n^{-4})^{-1}(x^{3})^{-1}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(n)^{-1}}}{{(n^{-4}x^{3})^{-1}}} = \dfrac{{n^{-1}}}{{n^{4}x^{-3}}}$ Break up the equation by variable and simplify. $\dfrac{{n^{-1}}}{{n^{4}x^{-3}}} = \dfrac{{n^{-1}}}{{n^{4}}} \cdot \dfrac{{1}}{{x^{-3}}} = n^{{-1} - {4}} \cdot x^{- {(-3)}} = n^{-5}x^{3}$.